Posted by SUPER HARD on .
In triangle ABC, m<A = 90, m<B=75, and BC = \sqrt{3}. What is the area of ABC? Express your answer as a common fraction.
HELP THIS IS SO HARD HHELP MEE!!! EXplain too please!

HELP ME HELP ME GEOMETRY PROBLM!!! 
Reiny,
You have a rightangled triangle so we can use the formula
(1/2) base x height
Since you want the answer as a common fraction, I will assume that you need "exact" values.
you state that BC = \sqrt{3}
I will assume that is meant to say √3
so cos75 =base/√3
base = √3cos75
and sin75 = height/√3
height = √3sin75
area = (1/2)(√3sin75)(√3cos75)
= (3/2)sin75cos75)
(recall sin 2A = 2sinAcosA)
= (3/2)(2sin75cos75)/2
= (3/4) sin 150
sin 150 = sin30  one of the basic identities
area = (1/3) sin150
= (3/4)sin30
= (3/4)(1/2) = 3/8 
HELP ME HELP ME GEOMETRY PROBLM!!! 
SUPER HARD,
thanks alot

HELP ME HELP ME GEOMETRY PROBLM!!! 
BOB,
THANK YOU !!!!!!!

HELP ME HELP ME GEOMETRY PROBLM!!! 
Ok OK,
You don't even need trig for this. Reflect B, draw and altitude from B to B'C and solve